Answer:
3242
Step-by-step explanation:
6
Answer:
C. Test for Goodness-of-fit.
Step-by-step explanation:
C. Test for Goodness-of-fit would be most appropriate for the given situation.
A. Test Of Homogeneity.
The value of q is large when the sample variances differ greatly and is zero when all variances are zero . Sample variances do not differ greatly in the given question.
B. Test for Independence.
The chi square is used to test the hypothesis about the independence of two variables each of which is classified into number of attributes. They are not classified into attributes.
C. Test for Goodness-of-fit.
The chi square test is applicable when the cell probabilities depend upon unknown parameters provided that the unknown parameters are replaced with their estimates and provided that one degree of freedom is deducted for each parameter estimated.
When I factor[ 2cos (square) - 5cos -3], I get (2cos + 1)(cos - 3). 2cos + 1 = 0, 2cos = -1, cos = -0.5,. Using inv cos on calculator, I get 120 degree related angle.
Answer:
Step-by-step explanation:
3/4 + (1/4 - 1/6)
= 3/4 +1/4 - 1/6
The LCM = 12
= {(3×3) + (1×3) - (2×1)}/12
= (9+3-2)/12
= 10/12
= 5/6
Answer:
(3,5)
∧
this is x number
Step-by-step explanation:
the x number is left and right movement. if you move 3 units left you would subtract 3 from the x number. so 6-3=3 and you don't have to do anything on the y value because you don't move it.
